Numerical Problems on Electromagnetic Induction (EMI)

Science > Physics > Electromagnetic Induction > Numerical Problems on Electromagnetic Induction

In this article, we shall learn to solve numerical problems to find e.m.f. induced, change in flux, and change in magnetic induction.

Example 01:

A circular coil having 1000 insulated turns each of area 1.274 m² is kept in a uniform magnetic field of induction 5 x 10^-4 T. If the plane of the coil makes an angle of 60^o with the direction of the magnetic field, find magnetic flux flowing through the coil.

Given: Number of turns of coil = n = 1000, Area of coil = A = 1.274 m², Magnetic induction = B = 5 x 10^-4 T,Angle between plane of coil and magnetic field = θ = 60^o.

To Find: Magnetic Flux = Φ =?

Solution:

Magnetic flux Φ = nAB cos θ

∴ Φ = 1000 x 1.274 x 5 x 10^-4 x cos 60^o

∴ Φ = 0.55 Wb

Ans: Magnetic flux flowing through the coil = 0.55 Wb

Example – 02:

The magnetic flux associated with a coil changes from zero to 6 x 10^-2 Wb in 0.6 s. Find the average e.m.f. induced in the coil

Given: Initial flux Φ₁ = 0, Final flux Φ₂ = 6 x 10^-2 Wb, Time in which change took place = dt = 0.6 s.

To Find: e.m.f. induced = e =?

Solution:

Change in flux dΦ = Φ₂ – Φ₁ = 6 x 10^-2 – 0

∴ dΦ = 6 x 10^-2 Wb

By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by

e = dΦ/dt = 6 x 10^-2 /0.6

∴ e = 0.1 V

Ans: e.m.f. induced = 0.1 V

Example – 03:

A coil of effective area 4 m² is placed at right angles to a magnetic field of induction 0.05 Wb m^-2. If the field is decreased by 20% of its original value in 10 seconds, find the e.m.f. induced in the coil

Given: Effective area of coil = nA = 4m², Angle between plane of coil and the magnetic field = θ = 90^o, Initial magnetic induction = B₁ = 0.05 Wb m^-2, Decrease in magnetic induction = 20%, Time in which change took place = dt = 10 s.

To Find: e.m.f. induced = e =?

Solution:

Final magnetic induction = B₂ = B₁ – 20% of B₁

∴ B₂ = 80% B₁ = 0.80 x 0.05 = 0.04 Wb m^-2

Initial flux Φ₁ = nAB₁ cos θ

∴ Φ₁ = 4 x 0.05 x cos 90^o = 0.2 Wb

Final Flux Φ₂ = nAB₁ cos θ

∴ Φ₂ = 4 x 0.04 x cos 90^o = 0.16 Wb

Change in flux dΦ= Φ₁ – Φ₂ = 0.2 – 0.16

∴ dΦ = 0.04 Wb

By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by

e = dΦ/dt = 0.04 /10 = 0.004 V

∴ e = 4 x 10^-3 V = 4 mV

Ans: e.m.f. induced = 0.4 mV

Example – 04:

A coil of effective area 4 m² is placed at right angles to a magnetic field of induction 0.05 Wb m^-2. If the field is decreased to 20% of its original value in 10 seconds, find the e.m.f. induced in the coil

Given: Effective area of coil = nA = 4m², Angle between plane of coil and the magnetic field = θ = 90^o, Initial magnetic field = B₁ = 0.05 Wb m^-2, New magnetic induction (B₂) = 20% of original induction (B₁), Time in which change took place = dt = 10 s.

To Find: e.m.f. induced = e =?

Solution:

Final magnetic induction = B₂ = 20% of B₁ = 0.20 x 0.05 = 0.01 Wb m^-2

Initial flux Φ₁ = nAB₁ cos θ

∴ Φ₁ = 4 x 0.05 x cos 90^o = 0.2 Wb

Final Flux Φ₂ = nAB₁ cos θ

∴ Φ₂ = 4 x 0.01 x cos 90^o = 0.04 Wb

Change in flux dΦ = Φ₁ – Φ₂ = 0.2 – 0.04

∴ dΦ = 0.16 Wb

By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by

e = dΦ/dt = 0.16 /10 = 0.016 V

∴ e = 16 x 10^-3 V = 16 mV

Ans: e.m.f. induced = 16 mV

Example – 05:

A coil of effective area 2 m² is placed at right angles to a magnetic field of induction 0.08 Wb m^-2. If the field is decreased to 10% of its original value in 0.6 seconds, find the e.m.f. induced in the coil

Given: Effective area of coil = nA = 2m², Angle between plane of coil and the magnetic field = θ = 90^o, Initial magnetic field = B₁ = 0.08 Wb m^-2, New magnetic induction (B₂) = 10% of original induction (B₁), Time in which change took place = dt = 0.6 s.

To Find: e.m.f. induced = e =?

Solution:

Final magnetic induction = B₂ = 10% of B₁

= 0.10 x 0.08 = 0.008 Wb m^-2

Initial flux Φ₁ = nAB₁ cos θ

∴ Φ₁ = 2 x 0.08 x cos 90^o = 0.16 Wb

Final Flux Φ₂ = nAB₁ cos θ

∴ Φ₂ = 2 x 0.008 x cos 90^o = 0.016 Wb

Change in flux dΦ = Φ₁ – Φ₂ = 0.16 – 0.016

∴ dΦ = 0.144 Wb

By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by

e = dΦ/dt = 0.144 /0.6

∴ e = 0.24 V

Ans: e.m.f. induced = 0.24 V

Example – 06:

A coil of 50 turns each of area 800 cm² is held with its plane perpendicular to a uniform magnetic field of induction 5 x 10^-5 T. If it is pulled out of the field in 2 s, find the e.m.f. induced in the coil.

Given: Number of turns of coil = n = 50, area of each turn of coil = A = 800m²= 800 x 10^-4 m², Angle between plane of coil and the magnetic field = θ = 90^o, Initial magnetic field = B₁ = 5 x 10^-5 T, New magnetic induction = B₂ = 0, Time in which change took place = dt = 2 s.

To Find: e.m.f. induced = e =?

Solution:

Initial flux Φ₁ = nAB₁ cos θ

∴ Φ₁ = 50 x 800 x 10^-4 x 5 x 10^-5 x cos 90^o

∴ Φ₁ = 2 x 10^-4 Wb

Final Flux Φ₂ = nAB₁ cos θ

∴ Φ₂= 50 x 800 x 10^-4 x 0 x cos 90^o

∴ Φ₂= 0 Wb

Change in flux dΦ = Φ₁ – Φ₂ = 2 x 10^-4 – 0

∴ dΦ = 2 x 10^-4 Wb

By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by

e = dΦ/dt = 2 x 10^-4 /2

∴ e = 1 x 10^-4 V

Ans: e.m.f. induced = 10^-4 V

Example – 07:

A rectangular coil of length 0.5 m and breadth 0.4 m has a resistance of 5 ohm is placed with its plane perpendicular to a uniform magnetic field of induction 0.05 T. The magnetic induction is uniformly reduced to zero in 5 x 10^-3 s. What is the e.m.f. induced and the current induced in the coil?

Given: Length of coil = l = 0.5 m, Breadth of coil = b = 0.4 m, Resistance of coil = R = 5 ohm, Angle between plane of coil and the magnetic field = θ = 90^o, Initial magnetic induction = B₁ = 0.05 T, Final magnetic induction = B₂ = 0, Time in which change took place = dt = 5 x 10^-3s.

To Find: e.m.f. induced = e =?, Current induced = I =?

Solution:

Area of the coil = A = l x b = 0.5 x 0.4 = 0.2 m²

Assuming the coil has single turn i.e. n = 1

Initial flux Φ₁ = nAB₁ cos θ

∴ Φ₁ = 1 x 0.2 x 0.05 x cos 90^o = 0. 0.01 Wb

Final Flux Φ₂ = nAB₁ cos θ

∴ Φ₂ = 1 x 0.2 x 0 x cos 90^o = 0 Wb

Change in flux dΦ = Φ₁ – Φ₂ = 0.01 – 0 6 = 0.01 Wb

By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by

e = dΦ/dt = 0.01 /5 x 10^-3

∴ e = 2 V

By Ohm’s law e = IR

I = e/R = 2/5 = 0.4 A

Ans: e.m.f. induced in coil = 2 V and current induced in coil = 0.4 A

Example – 08:

A coil of 50 turns each of length 0.5 m, and breadth 0.1 m is placed with its plane perpendicular to a uniform magnetic field of induction 10^-3 T. The magnetic induction is uniformly reduced to zero in 10^-2 s. What is the current induced in the coil if its resistance is 5 ohm.

Given: Number of turns of coil = n = 50, Length of coil = l = 0.5 m, Breadth of coil = b = 0.1 m, Resistance of coil = R = 5 ohm, Angle between plane of coil and the magnetic field = θ = 90^o, Initial magnetic induction = B₁ = 10^-3 T, Final magnetic induction = B₂ = 0, Time in which change took place = dt = 10^-2s.

To Find: Current induced = I =?

Solution:

Area of the coil = A = l x b = 0.5 x 0.1 = 0.05 m²

Assuming the coil has single turn i.e. n = 50

Initial flux Φ₁ = nAB₁ cos θ

∴ Φ₁ = 50 x 0.05 x 10^-3 x cos 90^o = 2.5 x 10^-3 Wb

Final Flux Φ₂ = nAB₁ cos θ

∴ Φ₂ = 50 x 0.05 x 0 x cos 90^o = 0 Wb

Change in flux dΦ = Φ₁ – Φ₂ = 2.5 x 10^-3 – 0 6 = 2.5 x 10^-3 Wb

By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by

e = dΦ/dt = 2.5 x 10^-3 / 10^-2

∴ e = 0.25 V

By Ohm’s law e = IR

I = e/R = 0.25/5 = 0.05 A

Ans: The current induced in coil = 0.05 A

Example – 09:

A rectangular loop having length 0.1 m and breadth 0.2 m and resistance 8 Ω is kept perpendicular to  a magnetic field, what change in the magnetic induction should be brought about in 0.5 s so as to induce a current of 4 mA in it.

Given: For a loop = n = 1, Length of coil = l = 0.1 m, Breadth of coil = b = 0.2 m, Resistance of coil = R = 8 Ω Angle between plane of coil and the magnetic field = θ = 90^o, Time in which change took place = dt = 0.5s, I = 4 mA = 4 x 10^-3 A.

To Find: Change in magnetic induction = dB =?

Solution:

Area of the coil = A = l x b = 0.1 x 0.2 = 0.02 m²

By Ohm’s law e = IR = 4 x 10^-3 x 8 = 0.032 V

By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by

e = dΦ/dt

∴ dΦ = e. dt = 0.032 x 0.5 = 0.016 Wb

∴ dΦ = f₁ – f₂ = nAB₁ cos θ – nAB₂ cos θ

∴ dΦ = nA(B₁ –B₂) cos θ

∴ dΦ = nA dB cos θ

∴ dB = dΦ/ nA cos θ = 0.016 / (1 x 0.02 x cos 90^o)

∴ dB= 0.8 T

Ans: Change in magnetic induction = 0.8 T

Example – 10:

A coil having an area of 0.5 m² and 1000 turns is kept perpendicular to a magnetic field of 0.05 T. Calculate the change in magnetic flux through the coil when it is rotated through an angle of 60^o about its diameter in 5 s. Find the e.m.f. induced in the coil.

Given: Number of turns = n = 1000, Area of coil = A = 0.5 m², Magnetic Induction = B = 0.05 T, Initial angle between plane of coil and the magnetic field = θ₁ = 90^o, New angle between plane of coil and the magnetic field = θ₂ = 60^o, Time in which change took place = dt = 5s.

To Find: e.m.f. induced = e =?

Solution:

Initial flux Φ₁ = nAB cos θ₁

∴ Φ₁ = 1000 x 0.5 x 0.05 x cos 90^o = 25 Wb

Final Flux Φ₂ = nAB cos θ₂

∴ Φ₂ = 1000 x 0.5 x 0.05 x cos 60^o = 12.5 Wb

Change in flux dΦ = Φ₁ – Φ₂ = 25 – 12.5 = 12.5 Wb

By the Faraday’s law of electromagnetism, the magnitude of induced e.m.f. is given by

e = dΦ/dt = 12.5 /5

∴ e = 2.5 V

Ans: Induced e.m.f. = 2.5 V